Published by:
CGP EDU Academic Team
Published on: August 14, 2026
For the reaction 
N 2 (g) + 
H 2 (g) ⎯→ NH 3 (g) ; Δ H = – 30 kJ to be at equilibrium at 477ºC. If standard entropy of N 2 (g) and NH 3 (g) are 60 and 50 J mole –1 K –1 respectively then calculate the standard entropy of H 2 (g) in Jmole –1 K –1 .
Text Solution
Verified by ExpertsThe correct answer is:
40
(40)
N 2 (g) + 
H 2 (g) ⎯→ NH 3 (g)
Δ S rxn = 50 – 30 – 
X JK –1
Δ H rxn = – 30000 J
at equilibrium
Δ G rxn = 0 = Δ H rxn – T Δ S rxn
750 × 
= – 30000
x = 
= 40.
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